Correct Answer - `-100^(@)C//m , (ii) 1000J`
(i) Temperature gradient = `(T_(1) - T_(2))/(l) = (100-0)/(1) = 100^(@)C//m`
(ii) Steady state temperature of element dx: `T = 100(1-x)`
Heat absorbed by the element to reach steady state. `dQ = "(dm)s" DeltaT = (gammadx)s(T-0)`
`implies dQ = 20[100(1-x)]dx`
Total heat absorbed by the rod `Q= intdQ = 2000 underset(0)overset(1)int(1-x)dx = 1000J`